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# Physics Problem

A uniform current carrying ring of mass $$m$$ and radius $$R$$ is connected by a massless string as shown. A uniform Magnetic field $$B_0$$ exist in the region to keep the ring in horizontal position, then the current in the ring is

$$(a)$$ $$\frac{mg}{\pi R B_0}$$

$$(b)$$ $$\frac{mg}{ R B_0}$$

$$(c)$$ $$\frac{mg}{3 \pi R B_0}$$

$$(d)$$ $$\frac{mg}{\pi R^2 B_0}$$

Note by Advitiya Brijesh
3 years, 8 months ago

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The tension of the string is $$T$$, and the weight force on the loop is $$mg$$. Because of the symmetry of the loop, the force caused by the magnetic field on the left half of the loop is equal and opposite to the force caused by the magnetic field on the right half. So the two net forces acting on the loop are $$T$$ and $$mg$$. The loop is at rest, so they must be equal.

$$T=mg$$.

The torque on the loop due to the string is $$RT$$, and the torque on the loop due to the magnetic field is $$B_0 A I$$, where $$A$$ is the area of the loop and $$I$$ is the current. The area of the loop is $$A=\pi R^2$$, so the torque due to the magnetic field is $$\pi R^2 B_0 I$$. The loop is at rest, so the two torques must be equal.

$$RT=\pi R^2 B_0 I$$

$$Rmg=\pi R^2 B_0 I$$

$$I=\frac{mg}{\pi R B_0} \quad (a)$$. · 3 years, 8 months ago

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T=mg (Force Balance).

TR=i$$\pi$$$$R^{2}$$$$B_{0}$$ (Torque Balance)

Solve the two to get answer as (a). · 3 years, 8 months ago

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Is ur Age 14 Mr. Advitiya B. ? · 3 years, 8 months ago

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Comment deleted Jul 25, 2014

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Did u Qualify in NTSE ? · 3 years, 8 months ago

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Comment deleted Jul 25, 2014

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I think the comment is... " I did'nt give that!!! :( "

I am curious ...becoz... My age person .in my Country.. . went to high Levels in challenges in Less TIME ... U joined from March 2013 right · 3 years, 8 months ago

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c) T + (2PIRBI) = Mg ....... BALANCING THE FORCES TR=(PIRRBi) ..................BALANCING TORQUE SOLVE TO GET i :) · 3 years, 8 months ago

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